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Question
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6 pF is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor.
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Solution
`E=1/2CV^2`
`E=1/2×12×10^-12×(50)^2`
`E=1.5×10^-8 J`
`C_(eqv)=(C_1C_2)/(C_1+C_2)`
`=(12×6)/(12+6)`
`=72/18`
`=4×10^-12 F`
`Q=C_(eqv)V`
`=4×10^-12×50`
`=200×10^-12 C`
Charge stored in both capacitor is `200×10^-12 C`.
Voltage across 12 pF capacitor is `=Q/C= (200×10^-12)/(12×10^-12)=200/12=16.67 V`
Voltage across 12 pF capacitor is`=Q/C= (200×10^-12)/(6×10^-12)=200/6=33.33 V`
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